Notesale: Turn your study into money

Document Preview

Extracts from the notes are below, to see the PDF you'll receive please use the links above

---

Created by Turbolearn AI

4
...
The study
of motion is heavily influenced by the choice of reference frame
...

In simpler terms, force can:
Change an object's velocity (accelerate or decelerate)
...

Force is a vector quantity, possessing both magnitude and direction
...
When multiple forces act on a body, the net force is the vector
addition of all forces
...


Weight
The weight of an object is the force with which the earth attracts that
object towards its center
...


Normal Force
When an object rests on a surface, such as a book on a table, it experiences a normal
force
...
If the contact surface is smooth, the direction of
the force is normal (perpendicular) to the plane of contact
...

Tension is a resisting force that is operative in a stretched string
...


Newton's First Law of Motion
A body observed from an inertial frame (at rest or moving with uniform
velocity) will remain at rest or continue to move with uniform velocity
unless an external force is applied
...
Newton's first law defines inertia, force, and inertial
frame of reference
...

Illustration 3: A heavy particle of mass 0
...
Find the force exerted by the string on the particle
...
8 m/s^2)

Page 2

Created by Turbolearn AI

Solution: The weight of the particle is balanced by the force of tension in the string
...
50kg ∗ 9
...
9N

, vertically downward (b) Pull of the string, T, vertically upward
...
Hence, the sum of forces should be zero
...
9 N when acting
vertically upward
...

Inertia is a passive property that resists active agents like torques and forces
...


Types of Inertia
There are three types of inertia:
Inertia of Rest: The inability of a body to change its state of rest by itself
...

Inertia of Motion: The inability of a body to change its state of uniform motion
by itself
...

Inertia of Direction: The inability of a body to change its direction of motion by
itself
...


Linear Momentum
Linear momentum measures an object's translational motion
...


Newton's Second Law of Motion
The net force on an object is equal to the rate of change of its linear
momentum in an inertial reference frame
...


Page 4

Created by Turbolearn AI

Illustration 5: Two forces
F1

and
F2

act on a 2 kg mass
...

Solution: Apply Newton's second law of motion
...

2
2
F = √ 10 + 5 + 2 ∗ 10 ∗ 5 ∗ cos(120) = √ 100 + 25 − 50 = √ 75 = 5√ 3N

5√ 3

F
a =

=
m

2

5

5 sin(120)
tan(θ) =

=
10 + 5 cos(120)

2
= 2
...
5√ 3m/sec

at an angle
30

∘

with the direction of
F1


...
Gravitational
2
...
Nuclear
4
...

The force of attraction between two masses is:
F = G

m1 m2
r2

Where:
and m = masses of the particles
r = distance between the particles
G = universal gravitational constant (6
...
85419 × 10
1

−12

9

C

2

2

/N m

= 9
...
Forces between surfaces in contact
2
...
Force due to a spring

Page 6

Created by Turbolearn AI

Nuclear Force
The nuclear force binds protons and neutrons in an atomic nucleus
...


Weak Force
The weak force underlies some forms of radioactivity
...
It acts upon fermions and is mediated by W and Z particles
...

Mathematically:
F AB = −F BA

The forces act on different bodies
...

(b) Second Law: Force = change of momentum with change of time, i
...
,
d(mv)
F =
dt

With mass constant: Force = mass x acceleration
F = ma

Force = mass x change in velocity with time
(V 1 − V 0 )
F = m
(t 1 − t 0 )

Hence, each has both magnitude and direction
...


Masterjee Concepts: Working with Laws of Motion
1
...

2
...

3
...

4
...

If forces are coplanar, use X and Y axes
...

Write equations by equating the sum of force components to the product of mass and
acceleration
...


Impulse
The impulse of a force is defined as the product of the average force
F

and the time interval
Δt

during which the force acts
...
The SI unit
is Newton-second (Ns)
...

F Δt = mv f − mv 0

Illustration 7: A truck of mass travelling at 4 m/s is brought to rest in 2 s when it
strikes a wall
...

Illustration 8: Assume that on a certain day rain comes down at a velocity of 15 m/s
and hits the roof of a car
...
060 kg/s
...

Solution:
Force on the roof of the car is equal to the momentum imparted to it per second by
rain drops
...
060 ∗ (0 − 15) = −0
...
If its speed also changes from 20 m/s to 10 m/s, then find the
magnitude of impulse acting on the bullet
...

Mass of the bullet,
m = 10

−3

kg

Consider components parallel to
J1

J 1 = 10

−3

∗ 10 ∗ cos(60) − (10

−3

∗ 20) = 10

−3

1
∗ 10 ∗

− 10

−3

∗ 20 = −15 ∗ 10

−3

N
...
s

2

The magnitude of the resultant impulse is given by $$J = \sqrt{J_1^2 + J_2^2} =
\sqrt{(-1510^{-3})^2 + (5\sqrt{3}10^{-3})^2} = \sqrt{22510^{-6} + 7510^{-6}} =
\sqrt{300*10^{-6}} = 10^{-3} * 10\sqrt{3} = 10^{-2} \sqrt{3} N
...

Lift Moving with Constant Velocity:
N = mg

The apparent weight equals the true weight
...

Lift Accelerating Downward with
a

:
mg − N = ma

N = m(g − a)

The apparent weight is lesser than the actual weight
...
At rest or constant velocity, these forces are
equal
...
At rest or constant velocity, these are equal
...


Page 13

Created by Turbolearn AI

Illustration 10: Two blocks of masses
M1

and
M2

are placed in contact with each other on a frictionless horizontal surface
...
Find the magnitude of acceleration of the system
...

Solution:
Applying Newton's second law for
M1

:
F1 − N = M1 a
N1 = M1 g

Applying Newton's second law for
M2

:
N − F2 = M2 a
N2 = M2 g

Page 14

Created by Turbolearn AI

Solving the equations,
F1 − F2

a =

M1 + M2
M2 F1 + M1 F2

N =

M1 + M2

Illustration 11: A rope of length L is pulled by a constant force F
...
30
...

Now, find: (a) The time required for the block A to move 1 m down the plane, starting
from rest; (b) The tension in the cord connecting the blocks
...
8 ∗
2

= 9
...
8 ∗

2
1

1 =

at

= 1
...
02 = 1
...
96

T = m B g − m B a = 20 ∗ 9
...
96 = 196 − 39
...
8N

Forces and Laws of Motion
Two-Block System on an Inclined Plane

Page 16

2

Created by Turbolearn AI

Consider two blocks, A and B, connected by a string, with block A on an
inclined plane
...

T represents the tension in the string
...
5m/s

∘

2

Law of Conservation of Linear Momentum
If no external forces act on a system of objects, the vector sum of the
linear momentum of each body remains constant
...
Define the system
...
Identify internal and external forces
...
Verify the system is isolated (no external forces)
...
Set final momentum equal to initial momentum
...

r

Find the recoil speed of the gun (v)
...

1

2

The man moves with velocity v relative to the platform
...

Momentum conservation:
0 = m 1 (v − v r ) + m 2 v v =

m1 vr
m 1 +m 2

Variable Mass Systems
Rocket Propulsion:
Thrust Force (F ) : F
t

Weight (W): W
Net Force (F

net

= vr

t

dm
dt

= mg

): F

= Ft − W

net

Net Acceleration (a): a =

dv
dt

=

vr

dm

m

dt

− g

Velocity at time t (v):
v = u − gt + v r ln

m0
m

Where:

is the exhaust velocity
m is the initial mass
m is the mass at time t
u is the initial velocity
vr

0

Falling Raindrop Example:

Page 18

Created by Turbolearn AI

A raindrop accumulates mass at a rate kmv (k > 0)
...

Equation of motion:
mg =

Given:
m

dv
dt

d
dt

(mv) = m

dm
dt

dv
dt

+ v

dm
dt

= kmv

= mg − kv

2

Equilibrium
Equilibrium is the condition where the net external force on a system is
zero
...

Dynamic equilibrium: Constant velocity (zero net force)
...


Constrained Motion
Motion of a body is controlled
Masses Connected by Pulley:
1
...

2
...

3
...

4
...

Constraint Relation:

Page 19

Created by Turbolearn AI

x 1 + x 2 + x 0 = L = constant
2

d x1
dt

2

2

+

d x2
dt

= 0

2

⟹

a 1 = −a 2

Tension and Acceleration:
a =
T =

m 1 −m 2
m 1 +m 2
2m 1 m 2
m 1 +m 2

g
g

Pseudo Forces
Fictitious force applied in Non-inertial Frames
...

Pseudo force: F

= −ma

(m = mass, a = acceleration of the frame)
Acts opposite to the acceleration of the frame
...


擦 Friction
Force opposing relative motion between two surfaces in contact
...

Kinetic friction (f ): Force that opposes motion of sliding surfaces
...

Acceleration towards the center
...


Radial and Tangential Acceleration

Page 21

Created by Turbolearn AI

Radial acceleration (a ) is given by:
r

ar =

v

2

r

where:
is the tangential speed
r is the radius of the circular path
v

Tangential acceleration (a ) is given by:
t

at = r

dv
dt

where:
dv
dt

is the rate of change of the tangential speed
...
The coefficient of friction between the road and the tire is μ
...
The net acceleration is the vector sum of centripetal acceleration and tangential
acceleration
...
The friction force provides the necessary force for this acceleration
...
The car will skid when the friction force reaches its maximum limit
...
The frictional force is μM g
...
The small mass m, in the horizontal plane, moves in a circular path
...

To find the frequency of rotation of the small mass m so that the large mass M
remains stationary:

Page 23

Created by Turbolearn AI

1
...
For mass m, the vertical component of tension balances its weight:
T cos(θ) = mg

3
...

4
...
Substitute T

= Mg

2

into the centripetal force equation:

M g sin(θ) = ml sin(θ)ω

2

6
...
The frequency of rotation f is:
f =

ω
2π

=

1
2π

√

Mg
ml

Centripetal Force
Centripetal Force: The force required to keep an object moving in a
circular path, directed towards the center of the circle
...
Consider a block of mass
m on a frictionless turntable connected to a string
...
This
observer introduces a fictitious outward force (centrifugal force) of magnitude

...

Coriolis Force: A pseudo force that acts on objects within a rotating frame,
perpendicular to both the velocity of the object and the axis of rotation
...
In
an inertial frame, no pseudo forces are required
...
A smooth groove AB of length L (≪ R) is
made on the surface
...

To find the time taken by the particle to reach point B:

Page 25

Created by Turbolearn AI

1
...

2
...

3
...


R

4
...
Using the equation of motion, s = ut +
L = 0 +

1
2

2

(ω R)t

1
2

at

2

, where s = L and u = 0:

2

6
...
The centripetal force is provided
in one or more ways:
Friction Only
Banking of Roads Only
Both Friction and Banking of Roads

Friction Only
If a car of mass m moves at a speed v in a horizontal circular arc of radius r, the
friction force (f ) provides the centripetal force:
f =

mv

2

r

The limiting value of friction is f
mv
r

≤ μN

, where N

2

≤ μmg

Page 26


...
By applying Newton's laws:
N sin(θ) =

mv

2

r

N cos(θ) = mg

Dividing the two equations:
tan(θ) =

v

2

rg

Therefore, the speed at which a car does not slide down even on a smooth track is:
v = √ rg tan(θ)

Friction and Banking of Roads
Three forces act on the vehicle:
Weight (mg)
Normal reaction (N )
Friction (f )
The magnitude of N and the direction/magnitude of f adjust so the resultant force is
towards the center
...

If v > √rg tan(θ), friction is inward
...

If v = √rg tan(θ), friction is zero
...
Find
the frictional force if the vehicle moves at:
(a) 5 m/s
(b) 15 m/s
Assume friction prevents slipping
...


(b) For v = 15m/s, friction acts downwards:
N sin(θ) + f cos(θ) =

mv

2

r

N cos(θ) − f sin(θ) = mg

Substituting values, we get f

= 2500√ 5, N

(downward)
...

The vertical component of tension balances the weight, and the horizontal
component supplies the centripetal force
...
In a rotor, a person
hangs against the wall without support due to the rotor's rotation
...

This provides the centripetal acceleration

...

mv

2

r

Therefore:
tan(θ) =

f
N

=

mv

2

mgr

=

v

2

rg

Centrifugal Force
Centrifugal Force: A pseudo force that appears to push an object away
from the center of a circular path when viewed from a non-inertial
(rotating) frame of reference
...

2

Illustration: Particle on a Rotating Table
A particle of mass m is placed on a horizontal circular table rotating with angular
velocity ω at a distance r from the center
...

Its speed is v when the string makes an angle θ with the vertical
...
A small ball inside
rotates with the bowl without slipping
...
At a place P on Earth:
Radius of rotation: r = R sin(ϕ), where R is the Earth's radius and ϕ is the
colatitude
...
Equator
A body weighs 98 N at the North Pole
...
At the North Pole: mg = 98N , so m =
2
...
ω =
= 7
...
R = 6400km
′

2π

98
9
...
27 × 10

−5

)

2

3

⋅ (6400 × 10 ) = 97
...
If
you cannot identify that object and the method of interaction (contact or field), then
the force DOES NOT EXIST!
Here are some commonly encountered forces:

绳 Ropes or Strings
Exert tension forces on the system
...

Directed along the direction of extension of the rope or string
...

Generally a push type of force directed toward the system
...

Frictional Force:
Parallel to the surface
...
e
...

Often assumed to be related to the normal force through a coefficient of
friction
...


力的 General Pushes or Pulls

Page 33

Created by Turbolearn AI

If a problem specifies that some object is being pushed or pulled in some
direction, assume that the force specified is being exerted by some physical
object
...

Often neglected because its effects are deemed negligible
...


Gravitational Force
Commonly called the weight of the system
...

Generally acts in the downward direction
...


Drawing a Free Body Diagram (FBD)
Step 1: Represent the system by a simple circle or square, focusing attention on the
forces on and the resulting acceleration of the system
...

Step 3: Include the acceleration vector as well, but distinguish it from the force
vectors by drawing it in a different way
...

The frictional force is directed opposite to the direction that the system slips
with respect to the surface
...

The weight is directed down
...


Picking a Coordinate System
Pick a coordinate system to determine the angles that the forces and accelerations
make with the coordinate axes
...

The answers you obtain thereafter must and will be independent of your choice
of coordinate system, but clever choices will help us to arrive at equations that
are more easy to solve
...

These two forces lie at the angle theta (given in the problem statement as 20
degrees) from one of the axis directions
...

F = ma

Step 2: Apply the basic equation to the problem
...
Recognize that every vector equation is
shorthand for two (or, more generally, three) scalar equations
...


Page 35

Created by Turbolearn AI

Examples:
x: n
y: n

x

+ f x + P x + W x = ma x

y

+ f y + P y + W y = ma y

Step 4: Determine what each component is in terms of the vector magnitude and
trigonometric functions of the associated angles
...

Substitute mg for W if you know the mass of the system
...

The frictional force is purely in the −x direction
...

The weight has negative x- and y-components
...

Therefore:
x: 0 + (−f ) + (+P cos θ) + (−mg sin θ) = m(+a)
y: (+n) + 0 + (−P sin θ) + (−mg cos θ) = m(0)
Step 5: Simplify the resulting equations and figure out where to go from here
...
Hence,

n

t

2
2
a = √a n + a t

Tangential component a is responsible for change of speed of a particle
...

In general, in any curvilinear motion, the direction of instantaneous velocity is
tangential to the path, while acceleration may assume any direction
...

t

t

n

dv
a t = component of  a along v = a cos θ =

an

dv
=

dt

= rate of change of speed
...

In a = a cos θ, if θ is acute, a will be positive, and speed increases
...
If θ is 90 , a is zero,
and speed will remain constant
...

Circular motion of decreasing speed: a is negative
...


= 0



---